The Monty Hall Problem Using Bayes’ Theorem
The Monty Hall Problem is perhaps the most famous mathematical brain teaser that involves conditional probabilities and the concept of variable change. In an issue of Parade magazine, one author gave a brief yet complete description of the problem that I consider the best. It goes as follows:
“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, ‘Do you want to pick door No. 2?’ Is it to your advantage to switch your choice?” (Parade Magazine)
So, should you switch? Initial intuition may lead you to believe that there is no advantage to switching versus not switching because, since there are two doors remaining, it seems as though there would be a 50% chance of winning regardless of which door you pick. However, this is not the case. Utilizing some of the information we learned about in class regarding conditional probabilities, we can use Bayes’ Rule to demonstrate that, in fact, these probabilities are not equal and there is a most optimal strategy in the Monty Hall Problem.
Let’s work through an example and create some equations to determine what the probability is for each outcome here and ultimately decide the best strategy:
P(Car@A) = P(Car@B) = P(Car@C)=1/3
Since each door is equally likely to contain the car in the beginning, let’s just say you randomly pick door A (this math will be the same regardless of the initial door selected). Immediately after this, Monty reveals to you that there is a goat behind door B. After getting this new information, you decide to compute some conditional probabilities to determine your best course of action.
Firstly, you want to know the probability that Monty opened door B given the car is behind door A. Since you chose door A, Monty could have opened either door B or door C.
P(Open B | Car@A) = 1/2
Secondly, you ask yourself what was the probability that Monty opened door B given the car is behind door B? Well, this one is simply 0 since we know Monty will not reveal the door with the car.
P(Open B | Car@B) = 0
Finally, you want to determine the probability that Monty opened door B given the car is behind door C. In this conditional probability, Monty has to open door B because he cannot open door A since it is your selection, and he cannot open door C since he will never reveal the car.
P(Open B | Car@C) = 1
Now that we know these conditional probabilities, we can combine them to determine the optimal outcome for each scenario after randomly selecting door A.
Chart of possible outcomes (source: http://ucanalytics.com/blogs/bayes-theorem-monty-hall-problem/)
Based on these computed conditional probabilities, it is clear that the optimal strategy is swapping doors in the Monty Hall problem as it has a ⅔ probability of resulting in winning as opposed to the ⅓ when you keep your original choice. Now, obviously this does not mean that you will win the single game that you are playing, but if you want to choose the strategy that would produce the highest probability of winning, then it is in your best interest to change doors when Monty asks.